I feel $\frac{\binom{5}{3}\binom{39}{10}}{\binom{44}{13}}$ is not the correct answer, correct answer should be $\frac{\binom{13}{3}\binom{39}{10}}{\sum_{x = 0}^{5}\left ( \binom{13}{x}\binom{39}{13-x} \right )}$ (that is $0.30$).
The only given thing is that $A$ and $B$ together have $8$ Spades and consequently $C$ and $D$ together will have $5$ Spades.
It is not given that which $8$ Spades out of the $13$ are contained by $A$ and $B$.
If it was given that which $8$ Spades are contained by $A$ and $B$ then $\frac{\binom{5}{3}\binom{39}{10}}{\binom{44}{13}}$ would have been the correct answer.
Explanation for $\frac{\binom{13}{3}\binom{39}{10}}{\sum_{x = 0}^{5}\left ( \binom{13}{x}\binom{39}{13-x} \right )}$ :
Favourable outcomes for choosing $C$ : Choosing any $3$ Spades from $13$ available Spades and choosing remaining $10$ cards from the $39$ cards would be a legitimate favourable choice that is we have $\binom{13}{3}\binom{39}{10}$ favourable choices.
All possible outcomes for choosing $C$ : As it is given $A$ and $B$ have $8$ Spades, it can be concluded that $C$ and $D$ together will have exactly $5$ Spades.
So number of Spades that may be contained by $C$ can be any element from the set $\left \{ 0, 1, 2, 3, 4, 5 \right \}$.
e.g. $C$ can have $5$ and $D$ can have $0$ Spades,
$C$ can have $4$ and $D$ can have $1$ Spades and so on.
The number of ways in which $C$ can have $x$ Spades is same as number of ways of choosing $x$ Spades from $13$ and number of ways of choosing $\left ( 13-x \right )$ other cards from $39$, that is $\binom{13}{x}\binom{39}{13-x}$.
$x$ can be any element of the set $\left \{ 0, 1, 2, 3, 4, 5 \right \}$.
So all possible outcomes $= \sum_{x = 0}^{5}\left ( \binom{13}{x}\binom{39}{13-x} \right )$.
Hence the desired Probability $= \frac{\binom{13}{3}\binom{39}{10}}{\sum_{x = 0}^{5}\left ( \binom{13}{x}\binom{39}{13-x} \right ) }\approx 0.30$.
